Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Factoring (Again)

  1. Mar 2, 2008 #1
    1. The problem statement, all variables and given/known data
    I had a topic somewhere about factoring, and now I have some more factoring problems I don't understand.

    such as:

    1)(x-1)[tex]^{3}[/tex] - (x+2)[tex]^{3}[/tex]

    2)64x[tex]^{3}[/tex] - 27y[tex]^{3}[/tex]

    3)3ab - 20cd -15ac + 4bd



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 2, 2008 #2
    1. If I use substitutions, it may make things a little easier.

    [tex]a=(x-1)[/tex]
    [tex]b=(x+2)[/tex]

    [tex]a^3-b^3=(a-b)(a^2+ab+b^2)[/tex]

    [tex](x-1)^3-(x+2)^3=[(x-1)-(x+2)][(x-1)^2+(x-1)(x+2)+(x+2)^2][/tex]
    [tex]=(x-1-x-2)(x^2-2x+1+x^2+x-2+x^2+4x+4)[/tex]

    Does this confuse you more? Continue simplifying and collecting like terms and it's solved.

    2. Can you re-write it?

    What number must you raise to the power of 3, to attain 64 and 27? You want to choose a number so you can raise both your coefficient and variable to the same power.

    3. Factor by grouping.
     
    Last edited: Mar 2, 2008
  4. Mar 3, 2008 #3
    No, that's not confusing, I understand what you're doing in 1.

    I know 4 cubed and 3 cubed = 64 and 27, but I don't know how it looks in factored form.
     
  5. Mar 3, 2008 #4

    HallsofIvy

    User Avatar
    Science Advisor

    Then you should understand that rocophyics just told you what it looks like in factored form: x3- y3= (x- y)(x2+ xy+ y2).
     
    Last edited by a moderator: Mar 3, 2008
  6. Mar 3, 2008 #5

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi Andy! :smile:

    64x[tex]^{3}[/tex] - 27y[tex]^{3}[/tex] = 4[tex]^{3}[/tex]x[tex]^{3}[/tex] - 3[tex]^{3}[/tex]y[tex]^{3}[/tex]: does that help?
     
  7. Mar 3, 2008 #6
    [tex]a^mb^m=(ab)^m[/tex]
     
  8. Mar 3, 2008 #7
    Oh, okay, so 2 is like 1, but with coefficients.
     
  9. Mar 3, 2008 #8

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    You've got 1) and 2) now.

    Have you got 3)?

    If not, there are various ways of doing it - one is to write it as a 2x2 matrix.

    [size=-2](if you're ok now, don't forget to mark thread "solved"!)[/size]​
     
  10. Mar 3, 2008 #9
    Yeah, I got 3,

    I believe this is right (3a + 4d)(b - 5c).

    I understand the formula for x^3 - y^3, but what the minus sign is instead a plus sign?
     
    Last edited: Mar 3, 2008
  11. Mar 3, 2008 #10
    [tex](x+y)(x^2-xy+y^2)[/tex]

    Replace y, with (-y) and tell me what your new equation is.
     
  12. Mar 4, 2008 #11

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    factoring x^3 + y^3 ?

    Yay! :smile:

    Hint: divide by y^3, so you get (x/y)^3 + 1.

    Put x/y = z, so you get z^3 + 1.

    Can you see how to factor that? :smile:

    (if not, come back for another hint)
     
  13. Mar 4, 2008 #12

    HallsofIvy

    User Avatar
    Science Advisor

    [tex]x^n- y^n= (x- y)(x^{n-1}+ x^{n-2}y+ x^{n-3}y^2\cdot\cdot\cdot+ x^2y^{n-3}+ xy^{n-2}+ y^{n-1}[/tex]
    for n any positive integer.

    [tex]x^n+ y^n= (x+ y)(x^{n-1}- x^{n-2}y+ x^{n-2}y^2\cdot\cdot\cdot- x^2y^{n-3}+ xy^{n-2}- y^{n-1}[/tex]
    for n any odd integer.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook